Proof of the Arithmetic Mean - Geometric Mean Inequality

To outline the proof, in the forward argument, we will show that the statement is true for larger and larger values of n nn (specifically for all nnn powers of 222). In the backward argument, we will show that if the statement is true for nnn variables, then it is also true for n−1 n-1n−1 variables. By combining these arguments, we show that the statement is true for any value of n nn, by first applying the forward argument to show that it is true for a power of 2 that is larger than nnn, and then using the backward argument to show that it is true for n nn.

Proof: (Forward) We will show this by induction. The base case 21=2 2^1=221=2 is proved in Complete the Square. For the induction step, suppose the statement is true for some 2k 2^k2k; we would like to show that the statement is true for 2k+1 2^{k+1}2k+1. Given {ai}i=12k+1 \{a_i\}_{i=1} ^{2^{k+1}}{ai​}i=12k+1​ positive real values, we divide the set in half (obtaining {ai}i=12k \{a_i\}_{i=1} ^{2^{k}}{ai​}i=12k​ and {ai}i=2k+12k+1 \{a_i\}_{i=2^k+1} ^{2^{k+1}}{ai​}i=2k+12k+1​), and then apply the induction hypothesis to each set.

The first inequality follows from using the Induction Hypothesis twice, while the second inequality follows from the 2-variable case, by setting x1=a1⋅a2⋯a2k x_1 = a_1 \cdot a_2 \cdots a_{2^k}x1​=a1​⋅a2​⋯a2k​ and x2=a2k+1⋅a2k+2⋯a2k+1 x_2 = a_{2^k+1} \cdot a_{2^k+2} \cdots a_{2^{k+1}} x2​=a2k+1​⋅a2k+2​⋯a2k+1​. This completes the argument for the forward step.

(Backward) We will now show that if the statement is true for k kk, then it is also true for k−1 k-1k−1. Assume that the statement is true for any set of k kk positive real values, i.e. that

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